Đáp án:

a,

$n_{AgCl}=\dfrac{22,96}{143,5}=0,16$ mol 

Đặt CTTQ alkane $A$ là $C_nH_{2n+2}$

$C_nH_{2n+2}+(2n+2)Cl_2\to C_nCl_{2n+2}+(2n+2)HCl$

$AgNO_3+HCl\to AgCl+HNO_3$

$ n_{HCl}=n_{AgCl}=0,16$ mol 

$\Rightarrow n_A=\dfrac{0,16}{2n+2}=0,02$ 

$\Rightarrow n=3$

Vậy CTPT $A$ là $C_3H_8$

CTCT: $CH_3-CH_2-CH_3$

b,

$C_3H_8+5O_2\xrightarrow{{t^o}} 3CO_2+4H_2O$ 

$\Rightarrow n_{O_2}=5n_A=0,02.5=0,1$ mol 

$\Rightarrow V_{O_2}=0,1.24,79=2,479L$ 

Vậy $V_{kk}=5V_{O_2}=2,479.5=12,395L$

`a)`

CTPT của A là: `C_nH_{2n+2}`

`C_nH_{2n+2} + (2n + 2)Cl_2 -> C_nCl_{2n+2} + (2n + 2)HCl`

`HCl + AgNO_3 -> AgCl + HNO_3`

`n_{AgCl} = (22,96)/(143,5) = 0,16(mol)`

Theo PT: 

`n_{HCl} = n_{AgCl} =0,16(mol)`

`n_{HCl} = (2n + 2).n_A`

`=> 0,02(2n + 2) = 0,16`

`<=> n = 3`

`=> A: C_3H_8`

`CTCT: CH_3-CH_2-CH_3`

`b)`

$C_3H_8 + 5O_2 \xrightarrow{t^o} 3CO_2 + 4H_2O$
Theo PT: `n_{O_2} = 5n_{C_3H_8} = 0,1(mol)`

`=> n_{kk} = 5n_{O_2} = 0,5(mol)`

`=> V_{kk} = 0,5.24,79 = 12,395(l)`